Question: Find $\lim_{h\to 0}\dfrac{3\arccos\left(\dfrac{1}{2}+h\right)-3\arccos\left(\dfrac{1}{2}\right)}{h}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{2\sqrt{3}}{3}$ (Choice B) B $-2\sqrt{3}$ (Choice C) C $\sqrt{3}$ (Choice D) D The limit doesn't exist
The limit expression has the following form: $\lim_{h\to 0}\dfrac{f(t+h)-f(t)}{h}$ Therefore, it describes a derivative of a certain function $f$ at a certain $x$ -value $t$. Can you recognize what $f$ and $t$ are? Since the numerator is $3\arccos\left(\dfrac{1}{2}+h\right)-3\arccos\left(\dfrac{1}{2}\right)$, we can tell that the function is $f(x)=3\arccos(x)$ and the $x$ -value is $\dfrac{1}{2}$. In other words, the limit expression is equal to $f'\left(\dfrac{1}{2}\right)$ for $f(x)=3\arccos(x)$. Let's find $f'(x)$ : $f'(x)=3\cdot \dfrac{-1}{\sqrt{1-x^2}}$ Now let's evaluate $f'\left(\dfrac{1}{2}\right)$ : $\begin{aligned}f'\left(\dfrac{1}{2}\right)&=3\cdot \dfrac{-1}{\sqrt{1-\left(\dfrac12\right)^2}} \\\\ &=\dfrac{-3}{\sqrt{\dfrac{3}{4}}} \\\\ &=\dfrac{-3}{\left(\dfrac{\sqrt{3}}{2}\right)} \\\\ &=\dfrac{-6}{\sqrt{3}} \\\\ &=-2\sqrt{3}\end{aligned}$ In conclusion, $\lim_{h\to 0}\dfrac{3\arccos\left(\dfrac{1}{2}+h\right)-3\arccos\left(\dfrac{1}{2}\right)}{h}=-2\sqrt{3}$.